Additional note on the Stirling Asymptotic Series

A while ago somebody posted a question on Stirling's Formula, to which I
replied, and you then asked me if I knew what the exact formula for the
nth term of the Asymptotic series was.

Since then, I have done a bit of searching on the subject, and have
come up with some interesting knowledge.

First, the formula you have posted on your board:

x! = r(x+1) = root (2*pi*x) * (x/e)^x* [Asymptotic series]

is in fact not an equality. It is an approximation - but of a special
kind.

What is strange about an asymptotic series is that as an *infinite*
series (we let partial sum S(n) approach infinity), for a fixed value
of x, this sum will diverge to infinity, but if we take a finite partial
sum of that series, and let the value of x approach infinity, then it
will become fully accurate. This is the difference between asymptotic
series, and the kind of convergence which we think of for normal series.
Thus the infinite Stirling asymptotic series does not converge.

This makes the formula only good for approximation purposes - however,
the more terms we use, depending on the value of x we plug in, the more
precise the approximation - but there will come a certain point in the
series (this point depends on what value of x is chosen) in which further
terms, instead of becoming smaller and smaller will just get bigger and
bigger and bigger, thus the infinite partial sum will diverge in the end.

Also, I know how one can develop the terms of the series, though it
is indeed complicated.

We can use:

x! = root (2*pi*x) ((x/e)^x)e^(u(x))

And u(x) is the function (an infinite series):

inf B(2n)
sum -----------------------
n=1 2n(2n-1)*x^(2n-1)

where B(2n) are the even-ordered Bernoulli numbers.
Here is a simple analytic expression for the even ordered Bernoulli numbers: